An ac voltage V=220 (2×10^3 t) Volt is applied to a series LCR...

Given $V = 220 \sin(2 \times 10^3 t)$ V, the peak voltage is $V_m = 220$ V and the angular frequency is $\omega = 2 \times 10^3$ rad/s.

The inductive reactance is $X_L = \omega L = (2 \times 10^3) \times (10 \times 10^{-3}) = 20 \, \Omega$.

The capacitive reactance is $X_C = \dfrac{1}{\omega C} = \dfrac{1}{(2 \times 10^3) \times (25 \times 10^{-6})} = \dfrac{1}{50 \times 10^{-3}} = 20 \, \Omega$.

Since $X_L = X_C$, the circuit is in resonance. The impedance of the circuit is $Z = \sqrt{R^2 + (X_L - X_C)^2} = R = 100 \, \Omega$.

The current amplitude is $I_m = \dfrac{V_m}{Z} = \dfrac{220}{100} = 2.2$ A.

Answer: $2.2$ A